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Question 97 of 129

Q.Find the equation of the straight line which passes through the intersection of the straight lines 5x−6y=15x-6y=1 and 3x+2y+5=03x+2y+5=0 and is perpendicular to the straight line 3x−5y+11=03x-5y+11=0.

Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2018Subjective· 2mImportance★★★★★
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Writing the family of lines through the intersection point as (5x−6y−1)+λ(3x+2y+5)=0(5x-6y-1)+\lambda(3x+2y+5)=0 and enforcing perpendicularity to 3x−5y+11=03x-5y+11=0 gives λ=45\lambda=45, leading to the line 5x+3y+8=05x+3y+8=0.

The family of lines passing through the intersection of 5x−6y=15x-6y=1 and 3x+2y+5=03x+2y+5=0 is:

(5x−6y−1)+λ(3x+2y+5)=0(5x-6y-1)+\lambda(3x+2y+5)=0

(5+3λ)x+(2λ−6)y+(5λ−1)=0(5+3\lambda)x+(2\lambda-6)y+(5\lambda-1)=0

The slope of this line is −5+3λ2λ−6-\dfrac{5+3\lambda}{2\lambda-6}.

The given line 3x−5y+11=03x-5y+11=0 has slope 35\dfrac{3}{5}. For perpendicularity, the product of slopes must be −1-1:

−5+3λ2λ−6×35=−1-\dfrac{5+3\lambda}{2\lambda-6}\times\dfrac{3}{5} = -1

3(5+3λ)5(2λ−6)=1\dfrac{3(5+3\lambda)}{5(2\lambda-6)} = 1

15+9λ=10λ−3015+9\lambda = 10\lambda-30

15+30=10λ−9λ⇒λ=4515+30 = 10\lambda-9\lambda \Rightarrow \lambda=45

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